Explanation Given information: The function f ( x , y ) = 1 − x + y − 3 x 2 y . Calculation: Express the partial derivative of the f ( x , y ) with respect to x at the point ( x 0 , y 0 ) . f ( x 0 , y 0 ) = lim h → 0 f ( x 0 + h , y 0 ) − f ( x 0 , y 0 ) h (1) Substitute 1 for x 0 and 2 for y 0 in Equation (1). f ( x 0 , y 0 ) = lim h → 0 f ( 1 + h , 2 ) − f ( 1 , 2 ) h Find the partial derivative of the f ( x , y ) with respect to x at the point ( x 0 , y 0 ) . f x ( 1 , 2 ) = lim h → 0 f ( 1 + h , 2 ) − f ( 1 , 2 ) h = lim h → 0 f [ 1 − ( 1 + h ) + 2 − 3 ( 1 + h ) 2 ] 2 − f [ 1 − 1 + 2 − ( 3 ) ( 1 ) 2 ( 2 ) ] h = lim h → 0 f [ − h + 2 − 6 − 12 h − 6 h 2 ] − f [ − 4 ] h = lim h → 0 f [ − 6 h 2 − 13 h − 4 ] − f [ − 4 ] h = lim h → 0 − 6 h 2 − 13 h h = lim h → 0 ( − 6 h − 13 ) = − 13 Thus, the partial derivative of the function at point ( 1 , 2 ) x is ∂ f ∂ x = − 13 _

University Calculus: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780321999580

Author: Joel R. Hass, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 13.3, Problem 57E

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Find the partial derivative of the function at the point (1,2).

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Chapter 13.3, Problem 56E

Chapter 13.3, Problem 58E

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Chapter 13 Solutions

University Calculus: Early Transcendentals (3rd Edition)

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Find the partial derivative of the function at the point ( 1 , 2 ) .

Find the partial derivative of the function at the point (1,2).

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Chapter 13 Solutions